Elemental calcium is produced by the electrolysis of molten \(\mathrm{CaCl}_{2}\). (a) What mass of calcium can be produced by this process if a current of \(7.5 \times 10^{3} \mathrm{~A}\) is applied for \(48 \mathrm{~h}\) ? Assume that the electrolytic cell is \(68 \%\) efficient. (b) What is the minimum voltage needed to cause the electrolysis?

Understanding Faraday's laws of electrolysis is essential when exploring the electrochemical process, as in the electrolysis of calcium chloride (CaCl₂). Faraday's first law states that the amount of substance liberated or dissolved at an electrode during electrolysis is directly proportional to the total electric charge passed through the substance.

This can be mathematically expressed as:
\[ m = (Q \/ F) \times (M \/ n) \]
where \( m \) is the mass of the substance (in grams), \( Q \) is the charge (in coulombs), \( F \) is Faraday's constant (approximately 96485 C/mol), \( M \) is the molar mass of the substance (in g/mol), and \( n \) is the number of moles of electrons exchanged per mole of substance in the reaction.

Faraday's second law complements the first, asserting that when the same amount of electric charge is passed through different substances, the mass of each substance produced or consumed will be proportional to its equivalent weight (which is the molar mass divided by the valency).

Using these principles, the quantity of calcium produced during electrolysis can be calculated with precision; but remember, knowing the efficiency of the cell and the number of electrons (n) in the redox reaction are crucial for accurate calculations.

The efficiency of an electrolytic cell is a measure of how effectively it converts electrical energy into chemical change. This is particularly important when calculating the actual yield of a product, like calcium, from the electrolysis of calcium chloride.

An electrolytic cell's efficiency is influenced by factors such as the purity of the reactants, the cell design, the conductive properties of the electrolyte, and temperature.

In practical applications, an electrolysis process is never 100% efficient due to energy losses from heat, side reactions, and resistance within the cell. This is why the problem specifies a 68% efficiency. To account for this in our calculation, we multiply the theoretical yield of calcium by the efficiency:\[ \text{Actual Mass} = \text{Theoretical Mass} \times \text{Efficiency} \]
This gives us a realistic mass of calcium that can be obtained from the electrolytic process.

The standard reduction potential is a crucial concept in electrochemistry, as it provides a quantitative measure of the tendency of a chemical species to acquire electrons and be reduced.

Each redox reaction or half-reaction has an associated standard reduction potential, measured in volts (V), and it is taken under standard conditions (25°C, 1 atm, and 1 M concentration). These values are reflected in the standard reduction potential table.

In our exercise, the standard reduction potential for the reaction:\[ \mathrm{Ca}^{2+}(l) + 2 \mathrm{e}^- \rightarrow \mathrm{Ca}(s) \] is -2.87 V. However, during electrolysis, the reaction is non-spontaneous; electrons are forced into the system, requiring external energy (an applied voltage) to occur. By applying the minimum voltage that is equal but opposite in sign to the standard reduction potential, we ensure that electrolysis can proceed.

In practical terms, to achieve efficient electrolysis, the applied voltage must not only meet the minimum required to activate the reaction but also overcome any additional resistances in the cell (e.g., from the electrodes or the electrolyte itself). Thus, the actual voltage used might be higher than the calculated standard potential.